I don't think that's what the question meant. It was more like an all or nothing scenario, but what is the better choice as far as probabilities go.

This is the type of question I'd be somewhat scared of in an interview. It's not all that difficult with time, but with someone staring at you or you are on the spot, it tends to be a lot harder. I've had simple questions asked of me and sometimes I freeze up for a second with 5 thoughts in my head at the same time "What does he really mean here? Is this a trick question? What answer will impress him the most? How'd she look naked?" etc. I usually get through it, but I'd be lying if I said I never bombed an interview question.

The confusing part to me was that I didn't recognize that if you wait until the end you EARN $10, whereas if you bet before your expectation is to EARN can only be a maximum of $9 (b/c you have to pay the $1 to play). When I first read the problem it sounded to me like the maximum possible was only $9 earned regardless of the choice.

And this was the other thing I thought was possible when I read the problem. I assumed if you won a bet, you could keep playing until all of the marbles were gone, with the last marble always being a freeroll. If this was the case, there would clearly be a better strategy if the marbles fell in your favor at some point (if they never fall in your favor, then just wait until the end).

The very thing you quoted implies that you only get to bet once. Otherwise, you wouldn't necessarily be broke.

"However, you can also choose to not bet and not pay your 1$. If you do not bet, and the final marble drawn is red, you win 10$, if it is blue, you win 0$."

I have $1. If I bet and lose I now have $0 and can no longer bet because I am broke. If I bet and win I now have $10 and am not broke. Why can't I bet with that money? This is how investing would work in reality.

That's not how the marble futures market works. You can only make 1 investment per 100 draws. This is due to new SEC regulations.

this is a correct way of thinking about it. try and think of it in terms of expected profit (which is a function of expected revenue - expected cost). the part after is slightly incorrect. hint: it involves the one dollar.

this is incorrect. but i agree the question is very suspectly worded. this was how it was read to me verbatim however so i figured it might be fun for you to try and solve it. i will answer any questions i can just as he did answer my questions.

this is not the answer though.

think about extreme (but simple) cases, where for example there were 2 red marbles, 1 blue marble left.

you are correct that when you place a bet you can only win 9 plus your 1 back. aside from that i'm not sure i understand the question.

the answer is there are scenarios in which all blues are taken, or all reds are taken. but these can be accounted for with probability distributions. good work, this is how i envisioned it initially (what happens if 10 and 10 are taken?)

think of it as the sample size shrinks even further and your edge approaches would would be an "ideal" scenario.

in these sorts of logic problems i have noticed that its usually best to think about the extreme cases one way or another as the solutions.

this is the incorrect value, but i like your line of reasoning. expound on it!

this is in essence the solution to half of the problem which i found (along with the EV of bet/no-bet). the other half is calculating the expected profit from either solution (bet or no bet).

he wouldn't tell me what the extension of the solution was but i can tell you that the EV of profit in the scenario of not betting i got (in for example, 2red, 1 blue remaining) was $6.67(no bet) and $5.67 (bet). to do this i just made a probability distribution tree and calculated the possible scenarios:

bet no bet
.66(r)*9, .33(b)*-1 .66(r) [.5red, .5blue] .33(b) [red 1.0]

my theory is basically in line with bruce, in that the nearest you can approach the guaranteed 1$ would only equal it. (for example if all blues were taken).

even in this extreme scenario you would have an expected profit of (if betting) 9$, vs not betting (10$).

in addition, due to the random sampling, i theorize that it would hypothetically never be optimal to bet vs wait because you would expect if you ran this game millions of times for the distribution of the drawn marbles to approach .5r/.5b. in that scenario it is undoubtedly profitable to not bet.

in addition: to those who asked about whether you can bet once. bruce is correct, you can only bet once. i struggled with this (thinking it would be an investment, like soebody had said, where we could maximize a shot once and then continue to bet if that edge remained. however, it is as bruce called it regulated such that you can ONLY BET ONCE. this was again something i had to ask.

there was another qeustion fromt he interview concerning options that was fun that i might post here (right before this question)

thanks guys! let me know if you have any other different answers or see holes in my logic.

It's not incorrect. The EV for the game when you never bet is $5. That's trivial because the last marble has probability 1/2 of being red.

There will be times after some marbles are drawn that your EV for waiting will be greater than $5. It can be as much as $10 when only reds are left. But the EV of waiting will always be greater than the EV of betting because whatever the probability is that the next marble is red, the probability that the last marble is red is the same, and we don't have to bet $1 on the last marble. That's the key point, that the probability for the last marble and the next marble are the same. There are no calculations necessary to reach this conclusion. I've already given the complete argument.


No, the 60% solution is completely irrelevant.


Of course it's $6.67. That's because the probability that the last one is red is 2/3. In the original game, it's 1/2, so the EV is $5. You don't need a decision tree. You're missing the key.

Bruce, part of the solution was calculating the expected profit in these various strategies, hence the calculations.

Bruce is trying to tell you that the calculation of the EV is extremely simple, only depending on the probability that the last marble is red.

gotcha kk.

Why should losing necessarily mean you lose your chance at the $10 at the end. That chance does not require you to place a wager. Nowhere is it explicitly stated that you cannot continue. It only says you can no longer wager.


In more than one way, this problem is very poorly stated. A lot of people thought you could bet more than once including the OP at first.

I don't know what you people's problem is. Seems crystal clear to me.